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Katyanochek1 [597]
3 years ago
13

When using the inspection method, the number you would add to (and subtract from) the constant term of the numerator so the poly

nomial in the numerator will have (x+ 3) as a factor is
Mathematics
1 answer:
krok68 [10]3 years ago
6 0

Question:

Consider the expression \frac{x^2 + x -10}{x + 3}

When using the inspection method the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x + 3) as a factor is?

Answer:

The constant to add is 4

Step-by-step explanation:

Given

\frac{x^2 + x -10}{x + 3}

First, we need to get an expression that has x + 3 has its factor.

Represent this expression with: (x + 3)(x + k)

Expand

x^2 + 3x + kx + 3k

Group like terms

x^2 + (3 + k)x + 3k

Compare the above expression to: x^2 + x - 10

(3 + k)x = x

3k = -10

However, we only consider solving (3 + k)x = x for k

(3 + k)x = x

3 + k = 1

Subtract 3 from both sides

3 - 3 + k = 1 - 3

k = 1 - 3

k= -2

Substitute -2 for k in (x + 3)(x + k)

(x + 3)(x + k) = (x + 3)(x -2)

(x + 3)(x + k) = x^2 + 3x - 2x - 6

(x + 3)(x + k) = x^2 + x - 6

So, the expression that has a factor of x + 3 is x^2 + x - 6

To get the constant term to add/subtract, we have:

Constant = (x^2 + x - 6) - (x^2 + x - 10)

Open brackets

Constant = x^2 + x - 6 - x^2 - x + 10

Collect Like Terms

Constant = x^2 - x^2+ x   - x - 6+ 10

Constant =  - 6+ 10

Constant =  4

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